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2 years ago

Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. 590 hours 281 hours 291 hours 568 hour 204 hours

Physics

1 answer:

alexgriva [62]2 years ago

3 0

The half-life of a nuclide is 737 hrs

Let the initial mass of nuclide is N° = 100

The final mass of nuclide after 407 hours

= 100 - 38 = 62

The expression for decay constant is

λ = 1/t ln(N°/Nt)

Substitute the given values and calculate the decay constant as,

λ = 1/407hr ln(100/62)

λ = 9.4 x 10^-4/hr

Now, the half-life is calculated as,

t1/2 = 0.693/λ

= 0.693/9.4 x 10^-4/hr

t1/2 = 737 hrs

Hence, the half-life of a nuclide is 737 hrs

Learn more about Half-Life here:

brainly.com/question/25750315

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Answer:

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Explanation:

Given,

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So, angular speed at t = 2.95 s

$\omega = \alpha t = 0.812 \times 2.95 = 2.4\ rad/s$

Now, KE of the merry go round

$KE = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2}\times 88.66\times 2.4^2 = 255.34 J$

Hence, the Kinetic energy of the merry go round = 255.34 J

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Answer:

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